Help with an inclined plane problem

In summary, the designer is trying to figure out how to activate the adaptor without overbalancing it and causing it to slide. They've looked at friction and normal forces between the wall and the pin, and between the pin and the wedge, but they still don't have a solution. They need to introduce unknowns for the three normal forces in order to solve for theta.
  • #1
DudleySquires
4
0
Hello folks. After a while lurking I've finally got a problem that I need some help with!

1. Homework Statement

I'm designing an adaptor for a drinks optic and have translated one of the aspects of the design into the following problem. There is plenty of stuff on the net about simple inclined plane problems but not quite like this.

D8lTb1U.png


In this problem the blue wedge can only move horizontally, constrained by the fixed green floor. The orange pin (viewed here end on) can only move vertically. It doesn't roll and is constrained by the fixed green wall.

The idea of this mechanism (if you haven't already guessed) is that the pin is raised by pushing the wedge under it.

The pin is pressed down by force Fdown.
The wedge is moved horizontally by Fmove.
Ignore gravity.

I need to determine the angle theta in terms of Fdown, μa & μb at the threshold of moving.

As I say, lots of stuff on the net on inclined planes, friction / frictionless, accelerating / equilibrium, but none quite the same as this where the inclined plane is moving under the pin.

Homework Equations


At threshold of moving Fmove = Fdown.
ΣFx = ΣFy = 0
μwall-pin = μpin-wedge = μa.
μwedge-floor = μb.

The Attempt at a Solution


I've drawn that many FBDs that it's not worth uploading them all. Suffice to say I know about component forces but I just can't figure how to make everything balance.

Thank you please!
 
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  • #2
hi, I'm glad you're posting interesting problems!
uh, you should probably write down all the important forces, acting on the pin, and acting on the wedge. Then you can start thinking about balancing things afterwards.
So maybe start with the pin. Which frictional forces are acting on it, and which external forces are acting on it? Also remember the normal forces.
 
  • #3
DudleySquires said:
At threshold of moving Fmove = Fdown.
Why would that be? Those two forces act at right angles.
 
  • #4
haruspex said:
Why would that be? Those two forces act at right angles.

One of the design constraints is that the force to activate the adaptor Fmove must be <= the force without the adapter Fdown. So I'm trying to deal only with the limit case where they are the same.
 
  • #5
Here's where I've got to. I've got as far as balancing the pin, then the wedge, and now I'm considering how the friction & perp forces between wall and pin should be accounted for.

JuVtZeO.jpg


KfsY862.png


I've used a friction coefficient of 0.3, which has resulted in the "slip" beginning at ~17deg. When that happens, the wall begins to react some of the force, which also sets up a new perp and associated friction force between the wall and pin. I suppose that any perp force that cannot be reacted by the friction force will have to be reacted by the ground, which is in addition to the vertical force that is already present.
 
Last edited:
  • #6
Must admit to being surprised at having created a problem which it appears that nobody has tackled before. As I say I can find nothing on the net. Even if I modify the question to say something like find what value of Fmove is required to move the pin upwards, in terms of theta and the friction coefficients, there is still nothing out there that I can find.
 
  • #7
DudleySquires said:
One of the design constraints is that the force to activate the adaptor Fmove must be <= the force without the adapter Fdown. So I'm trying to deal only with the limit case where they are the same.
Ok, but then you should have written |Fmove | =|Fdown |, and it should have been as part of the problem description. "Relevant equations" is for standard equations, such as conservation laws.
You need to introduce unknowns for the three normal forces. You do not know in advance what these are. They're not simply components of the force needed to balance gravity. I would suggest not assuming |Fmove | =|Fdown | yet - you can plug that in at the end.
On the roller, you have six forces: gravity, the propulsive force, two normal forces and two frictional forces. Write out two equations containing them.
Likewise the wedge, with its five forces.
Since either nothing moves or sliding occurs at all surfaces, you can take the frictional forces to be at maximum everywhere.
That gives you four equations and four unknowns: the three normal forces and theta. Solve.
 

Related to Help with an inclined plane problem

1. How do I calculate the mechanical advantage of an inclined plane?

The mechanical advantage of an inclined plane is calculated by dividing the length of the ramp by the height of the ramp. This will give you the ratio of the effort force to the load force.

2. What is the formula for calculating the force required to move an object up an inclined plane?

The formula for calculating the force required to move an object up an inclined plane is force = mass x acceleration. In this case, the acceleration will be the gravitational acceleration, which is 9.8 m/s^2.

3. How does friction affect the movement of an object on an inclined plane?

Friction can affect the movement of an object on an inclined plane by creating a resistance force that opposes the motion of the object. This can make it more difficult to move the object up the ramp and can also cause the object to slow down or stop if the ramp is not steep enough.

4. Can an inclined plane have a mechanical advantage greater than 1?

Yes, an inclined plane can have a mechanical advantage greater than 1. This means that the effort force required to move the object up the ramp is less than the load force, making it easier to move the object.

5. How can I use an inclined plane to reduce the amount of force needed to lift an object?

An inclined plane can be used to reduce the amount of force needed to lift an object by increasing the distance over which the object is lifted. This allows for a smaller amount of force to be applied over a longer distance, making it easier to move the object.

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